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Q: What is the answer for x0 for the function y12x plus 88?

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The general equation for a linear approximation is f(x) ≈ f(x0) + f'(x0)(x-x0) where f(x0) is the value of the function at x0 and f'(x0) is the derivative at x0. This describes a tangent line used to approximate the function. In higher order functions, the same concept can be applied. f(x,y) ≈ f(x0,y0) + fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0) where f(x0,y0) is the value of the function at (x0,y0), fx(x0,y0) is the partial derivative with respect to x at (x0,y0), and fy(x0,y0) is the partial derivative with respect to y at (x0,y0). This describes a tangent plane used to approximate a surface.

let f be a function and f' the first derivative.If f'>0 the function is genuinely ascending.If f'

The answer depends on the nature of the function that defines the curve whose slope you want. If the function f(x) is differentiable, its slope is f'(x) = df(x)/dx and the value of the slope at a point when x = x0 is f'(x0), obtained by substituting x0 for x in f'(x).

When your input variable causes your denominator to equal zero. * * * * * A rational function of a variable, x is of the form f(x)/g(x), the ratio of two functions of x. Suppose g(x) has a zero at x = x0. That is, g(x0) = 0. If f(x0) is not also equal to 0 then at x = x0 the rational function would involve division by 0. But division by 0 is not defined. Depending on whether the signs of f(x) and g(x) are the same or different, as x approaches x0 the ratio become increasingly large, or small. These "infinitely" large or small values are the asymptotes of the rational function at x = x0. If f(x0) = 0, you may or may not have an asymptote - depending on the first derivatives of the two functions.

shut up and do your hw

It's a method used in Numerical Analysis to find increasingly more accurate solutions to the roots of an equation. x1 = x0 - f(x0)/f'(x0) where f'(x0) is the derivative of f(x0)

0! You said x0! anything x0=0!

For a quadratic function, there is one minimum/maximum (the proof requires calculus) and also it is either always convex or concave (prove is also calculus) it is continuous every where, hence, it can have a maximum of 2 roots. Graph it. If there is more than 2 roots, by Intermediate Value Theorem, it cannot be convex/concave everywhere. It will HAVE to have two intervals of increasing or decreasing. It can be easily proven that given any quadratic function f(x), if x = x0 is a minimum/maximum, and x=a != x0 is a root, then 2x0-a is also a root. It is still true that a = x0 as 2x0-x0=x0 implying it is the only root. But the concept of min/max requires Calculus to prove existence. So, this is Calculus, not algebra.

Anything multiplied by 0 is 0, so all the ones before the 0 can be ignored. It is now 0+1 which equals 1.

x0 = 1 because any number raised to the power of 0 is always equal to 1

On a transformer connection H1 and H2 are the primary connections. X1 and X2 are the secondary connections. If your transformer has a split secondary that is grounded, that terminal is X0. The sequence is X1 - X0 - X2. The X0 usually indicates that there is a connection to a neutral wire along with the ground wire.

Function f(x) x2=x*x f=x*x2-3*x2+2*x-1 return end function f1(x) f1=3*x*x-6*x+2 return end 5 write(*,*)'enter initial guess' read(*,*) x0 write(*,*)'enter tolerrence' read(*,*) eps 10 x1=x0-f(x0)/f1(x0) if(abs((x1-x0)/x0).lt.eps) then write(*,1) x1 1 format('solution=',f10.4) stop else x0=x1 go to 10 endif end

#include#include#include#includefloat eq(float);void falsi(float,float,int);void main(){float x0,x1;int iter;clrscr();cout

most commonly you would place the highest exponent first. x0 = 1. 54x3 + 62x2 - 344x + 2 = 0

A function f(x), of a variable x, is said to have a limiting value of f(xo) as x approaches x0 if, given any value of epsilon, however small, it is possible to find a value delta such that |f(x) - f(x0)| < epsilon for all x such that |x - x0| < delta.The second inequality can be one-sided.

The integral of e-2x is -1/2*e-2x + c but I am not sure what "for x0" in the question means.

Assuming you want the equation of the straight line between the two points (x0, y0) and (x1, y1), the equation is: y - y0 = m(x - x0) where m is the gradient between the two points: m = (y1 - y0) ÷ (x1 - x0) Note: if the two x coordinates are equal, that is x0 = x1, then the equation of the line is x = x0.

In order to get the results of 0x1*2-1*x0 you will have to do a little math. The answer to this math problem is X equals one.

The equation of a sphere with radius r, centered at (x0 ,y0 ,z0 ) is (x-x0 )+(y-y0 )+(z-z0 )=r2

a cool one x0

x2 + x1 + x0

Any number to the power zero is equal to one. That can be derived from the following index law: xa*xb = xa+b (x not zero) Now let b = 0 so that the above becomes xa*x0 = xa+0 so xa*x0 = xa (since a+0 = a) That is, any number multiplied by x0 is the number itself. That can be true only if x0 is the multiplicative identity, that is, only if x0 = 1.

Zero. x0 = 1.

the shore ! lol x0

Probably... What ratio? X/R? X0/X1? X0/R0? Xd /Xd''? etc.